Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        A Projectile thrown with the speed 100m/s. Making  an angle 60°. With the horizontal. Find the time after which the inclination is 45°. `
28 days ago

Vikas TU
9499 Points
```							Dear stuent Initial and final velocities along the horizontal direction and vertical direction are :ux = ucos60 uy = usin60 vx = vcos45 vy = vsin45 Using v = u + at ay = -9.8m/s^2 ax = 0 m/s^2 Vx = ux + axt v = u/√2 ........ (1)Along vertical direction Vy = uy + ayt v/√2 = u√3/2 - 9.8t Put this in equation 1 and find the vaue of t t = 3.73 sec h = uyt + 0.5ayt^2 h = 18.43 mHope this helpsGood Luck
```
28 days ago
Arun
23032 Points
```							 let after time t its inclination with horizontal is 45   at this time Vy /Vx =tan45  or Vy =Vx     Vx remains constant =100cos60 =Vy     now using equation of motion for verticle direction    Vy=Uy-gt..........eq1   Uy=100sin60 ,Uy=100cos60  putting in eq 1,     you can find t
```
28 days ago
Think You Can Provide A Better Answer ?

## Other Related Questions on General Physics

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 18 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions